Question

Prove that
2 sec²Θ - sec⁴Θ - 2 cosec²Θ + cosec⁴Θ = cot⁴Θ - tan⁴Θ

Answer 1

SOLUTION :

Given :

2sec²θ–sec⁴θ–2cosec²θ+cosec⁴θ = cot⁴θ–tan⁴θ

L.H.S = 2sec²θ–sec⁴θ–2cosec²θ+cosec⁴θ

= (cosec⁴θ - 2cosec²θ) - ( sec⁴θ - 2sec²θ)

= (cosec⁴θ - 2cosec²θ + 1) - (sec⁴θ - 2sec²θ +1)

= ((cosec²θ)² - 2cosec²θ + 1²) - (( sec²θ )² - 2sec²θ + 1²)

=( cosec²θ - 1)² - (sec²θ -1)²

[ a² + b² - 2ab = (a - b)²]

= (cot²θ)² - (tan²θ)²

[ cosec²θ- 1= cot²θ , sec²θ - 1 = tan²θ]

LHS = cot⁴θ - tan⁴θ = RHS

HOPE THIS ANSWER WILL HELP YOU.

Answer 2

To prove:

⇒ 2sec²θ - sec⁴θ - 2cosec²θ + cosec⁴θ = cot⁴θ - tan⁴θ

From the LHS:

⇒ 2sec²θ - sec⁴θ - 2cosec²θ + cosec⁴θ

→ Take sec²θ outside since it's common to both 2sec²θ - sec⁴θ.

→ Take cosec²θ outside since it's common to both - 2cosec²θ + cosec⁴θ.

⇒ sec²θ[2 - sec²θ] - cosec²θ[2 - cosec²θ]

→ Using sec²θ = 1 + tan²θ and cosec²θ = 1 + cot²θ we get:

⇒ (1 + tan²θ)[2 - (1 + tan²θ)] - (1 + cot²θ)[2 - (1 + cot²θ)]

⇒ (1 + tan²θ)[2 - 1 - tan²θ] - (1 + cot²θ)[2 - 1 - cot²θ]

⇒ (1 + tan²θ)[1 - tan²θ] - (1 + cot²θ)[1 - cot²θ]

Using a² - b² = (a + b)(a - b) we get:

⇒ (1)² - (tan²θ)² - [(1)² - (cot²θ)²]

⇒ 1 - tan⁴θ - [1 - cot⁴θ]

⇒ 1 - tan⁴θ - 1 + cot⁴θ

⇒ - tan⁴θ + cot⁴θ

Rearranging the terms we get:

⇒ cot⁴θ - tan⁴θ

LHS = RHS

Hence Proved.